Question

一压力锅上标有“XX铝制品厂24公分压力锅”的字样,小明测得限压阀的质量是100g,排气孔直径是3mm,

Answer 1

限压阀对锅的压力为

F=mg=100 g×9.8 N/kg

=0.1 kg×9.8 N/kg=0.98 N

受力面积为

S=πr2=3.14×(3×10-3 m/2)2

=7.065×10-6 m2

限压阀对水蒸气的压强为

p=F/S=0.98 N/(7.065×10-6 m2)

=1.39×105 Pa

锅内水蒸气的最大压强为

p+p0=1.013×105 Pa+1.39×105 Pa

=2.403×105 Pa≈2.4p0

锅盖要能承受的压力为

F=pS=2.403×105 Pa×πr2

=2.403×105 Pa×3.14×(24×10-2 m/2)2

=1.09×104 N